Solve y=-(x-16)²: Finding Values Where Function is Negative

Q: Why is the function negative everywhere except x = 16?

The negative sign in front of the squared term makes this parabola open downward. Since $ (x-16)^2 \geq 0 $ for all real numbers, multiplying by -1 gives $ -(x-16)^2 \leq 0 $. It only equals zero when x = 16!

Q: How do I know the vertex is at (16, 0)?

In vertex form $ y = a(x-h)^2 + k $, the vertex is at (h, k). Here we have $ y = -(x-16)^2 + 0 $, so the vertex is at (16, 0).

Q: What does 'downward-opening parabola' mean?

When the coefficient of the squared term is negative (like our -1), the parabola opens downward like an upside-down U. This means the vertex is the highest point, and all other points are below it.

Q: Can I test specific values to check my answer?

Absolutely! Try x = 0: \( y = -(0-16)^2 = -256 . Try x = 20: \( y = -(20-16)^2 = -16 . Both are negative, confirming our answer!

Q: Why isn't the answer 'all x values'?

Because when x = 16, we get $ y = -(16-16)^2 = 0 $. Since we need f(x) (strictly less than), we cannot include x = 16 where f(x) = 0.

Quadratic Functions with Vertex Form Analysis

Given the function:

$y=-\left(x-16\right)^2$

Determine for which values of X the following is true:

$f(x) < 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Given the function:

$y=-\left(x-16\right)^2$

Determine for which values of X the following is true:

$f(x) < 0$

Step-by-step solution

To solve this problem, we'll analyze the given quadratic function:

The function is $y = -\left(x - 16\right)^2$ , a downward-opening parabola with vertex at $(16, 0)$ .
The quadratic function is in vertex form $y = -\left(x - 16\right)^2$ . Here, the value of $y$ is zero when $x = 16$ .
Since $-\left(x - 16\right)^2$ represents a downward-opening parabola, $y$ will be zero only when $x = 16$ .
For $y$ to be less than zero, $x$ must be any real number except 16, as the squared term results in zero exactly when $x = 16$ .
Thus, the inequality $f(x) < 0$ holds for all $x$ except at $x = 16$ .

Therefore, the solution to the problem is $x\ne16$ .

Final Answer

$x\ne16$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: Function $y = -(x-16)^2$ has vertex at (16, 0)
Sign Analysis: Negative coefficient makes parabola open downward from vertex
Check: Test x = 10: $y = -(10-16)^2 = -36 < 0$ ✓

Common Mistakes

Avoid these frequent errors

Thinking the function is positive everywhere except x = 16
Don't assume negative coefficient means positive values = backwards thinking! The negative sign in front means the parabola opens downward, making y-values negative except at the vertex. Always remember that $-(anything)^2$ gives negative or zero values only.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of $$ x $$ where $f\left(x\right) > 0$ .

FAQ

Everything you need to know about this question

Why is the function negative everywhere except x = 16?

The negative sign in front of the squared term makes this parabola open downward. Since $(x-16)^2 \geq 0$ for all real numbers, multiplying by -1 gives $-(x-16)^2 \leq 0$ . It only equals zero when x = 16!

How do I know the vertex is at (16, 0)?

In vertex form $y = a(x-h)^2 + k$ , the vertex is at (h, k). Here we have $y = -(x-16)^2 + 0$ , so the vertex is at (16, 0).

What does 'downward-opening parabola' mean?

When the coefficient of the squared term is negative (like our -1), the parabola opens downward like an upside-down U. This means the vertex is the highest point, and all other points are below it.

Can I test specific values to check my answer?

Absolutely! Try x = 0: $y = -(0-16)^2 = -256 < 0$ . Try x = 20: $y = -(20-16)^2 = -16 < 0$ . Both are negative, confirming our answer!

Why isn't the answer 'all x values'?

Because when x = 16, we get $y = -(16-16)^2 = 0$ . Since we need f(x) < 0 (strictly less than), we cannot include x = 16 where f(x) = 0.

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Solve y=-(x-16)²: Finding Values Where Function is Negative

Quadratic Functions with Vertex Form Analysis

❤️ Continue Your Math Journey!

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why is the function negative everywhere except x = 16?

How do I know the vertex is at (16, 0)?

What does 'downward-opening parabola' mean?

Can I test specific values to check my answer?

Why isn't the answer 'all x values'?

🌟 Unlock Your Math Potential

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